Question: You have found the following ages (in years) of all 6 lions at your local zoo: $ 8,\enspace 4,\enspace 4,\enspace 1,\enspace 15,\enspace 2$ What is the average age of the lions at your zoo? What is the variance? You may round your answers to the nearest tenth.
Explanation: Because we have data for all 6 lions at the zoo, we are able to calculate the population mean $({\mu})$ and population variance $({\sigma^2})$ To find the population mean , add up the values of all $6$ ages and divide by $6$ $ {\mu} = \dfrac{\sum\limits_{i=1}^{{N}} x_i}{{N}} = \dfrac{\sum\limits_{i=1}^{{6}} x_i}{{6}} $ $ {\mu} = \dfrac{8 + 4 + 4 + 1 + 15 + 2}{{6}} = {5.7\text{ years old}} $ Find the squared deviations from the mean for each lion. Age $x_i$ Distance from the mean $(x_i - {\mu})$ $(x_i - {\mu})^2$ $8$ years $2.3$ years $5.29$ years $^2$ $4$ years $-1.7$ years $2.89$ years $^2$ $4$ years $-1.7$ years $2.89$ years $^2$ $1$ year $-4.7$ years $22.09$ years $^2$ $15$ years $9.3$ years $86.49$ years $^2$ $2$ years $-3.7$ years $13.69$ years $^2$ Because we used the population mean $({\mu})$ to compute the squared deviations from the mean , we can find the variance $({\sigma^2})$ , without introducing any bias, by simply averaging the squared deviations from the mean $ {\sigma^2} = \dfrac{\sum\limits_{i=1}^{{N}} (x_i - {\mu})^2}{{N}} $ $ {\sigma^2} = \dfrac{{5.29} + {2.89} + {2.89} + {22.09} + {86.49} + {13.69}} {{6}} $ $ {\sigma^2} = \dfrac{{133.34}}{{6}} = {22.22\text{ years}^2} $ The average lion at the zoo is 5.7 years old. The population variance is 22.22 years $^2$.